Knowing that $ i ^ 2 = -1 $, find all the actual $ x $ values that satisfy the following inequality: $$Re \left\{\frac{2 \log_2 \sin(x)+1}{i \left(e^{2ix}-2 \cos^2(x)+1 \right)}\right\}>0$$Where $ Re $ {$ Z $} is the real part of the $ Z $ complex number
I did the reduced form:
You are right, indeed we have that
$$\frac{1}{e^{2ix}-2 \cos^2(x)+1 }=\frac{1}{\cos(2x)-2 \cos^2(x)+1+i\sin (2x) }=\frac{1}{i\sin (2x) }$$
therefore
$$\frac{2 \log_2 \sin(x)+1}{i \left(e^{2ix}-2 \cos^2(x)+1 \right)}=-\frac{2 \log_2 \sin(x)+1}{\sin(2x)}$$
and we need to solve $\frac{2 \log_2 \sin(x)+1}{\sin(2x)}<0$.
Then let consider separately the numerator and the denominator, notably
$$\sin(2x)>0 \iff 2k\pi <2x<\pi+2k\pi $$
$$\sin(2x)<0 \iff \pi+ 2k\pi <2x<2\pi+2k\pi $$
and
$$2 \log_2 \sin(x)+1> 0 \iff \log_2 \sin(x)>-\frac12=\log_2 \frac1{\sqrt 2} \iff \sin x>\frac{\sqrt 2}2 $$$$\iff \frac \pi 4+ 2k\pi <x<\frac {3\pi} 4+2k\pi$$
$$2 \log_2 \sin(x)+1< 0 \iff \log_2 \sin(x)<-\frac12=\log_2 \frac1{\sqrt 2} \iff 0<\sin x<\frac{\sqrt 2}2$$$$\iff 2k\pi <x<\frac {\pi} 4+2k\pi \cup \frac {3\pi} 4+2k\pi <x<\pi+2k\pi$$
and then put togheter the intervals case by case.